Lemma 101.21.5. Let $\mathcal{X}$ be an algebraic stack. Consider a cartesian diagram

$\xymatrix{ U \ar[d] & F \ar[l]^ p \ar[d] \\ \mathcal{X} & \mathop{\mathrm{Spec}}(k) \ar[l] }$

where $U$ is an algebraic space, $k$ is a field, and $U \to \mathcal{X}$ is flat and locally of finite presentation. Let $z \in |F|$ be such that $F \to \mathop{\mathrm{Spec}}(k)$ is unramified at $z$. Then, after replacing $U$ by an open subspace containing $p(z)$, the morphism

$U \longrightarrow \mathcal{X}$

is étale.

Proof. Since $f : U \to \mathcal{X}$ is flat and locally of finite presentation there exists a maximal open $W(f) \subset U$ such that the restriction $f|_{W(f)} : W(f) \to \mathcal{X}$ is étale, see Properties of Stacks, Remark 100.9.20 (5). Hence all we need to do is prove that $p(z)$ is a point of $W(f)$. Moreover, the remark referenced above also shows the formation of $W(f)$ commutes with arbitrary base change by a morphism which is representable by algebraic spaces. Hence it suffices to show that the morphism $F \to \mathop{\mathrm{Spec}}(k)$ is étale at $z$. Since it is flat and locally of finite presentation as a base change of $U \to \mathcal{X}$ and since $F \to \mathop{\mathrm{Spec}}(k)$ is unramified at $z$ by assumption, this follows from Morphisms of Spaces, Lemma 67.39.12. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).